# finding length of airport runway

by NDbogan
Tags: distance, jumbo jet, runway length
 P: 18 the landing speed of a jumbo jet is 300Kn/hr. After landing, the jet comes to the safe taxiing speed of 30Km/hr in 12 seconds. The airport authorities would like to keep a saftey margin of 250 metres of runway. How long should the runway be for the safe landing of jumbo jets? Now, I remember doing this in class. I just can't find my notes. I was thinking that it might have something to do with: Final velocity(v)= initial velocity(u)+acceleration(a) x time(t) maybe re-arranging it then being able to find the distance travelled. so: a=(u-v)/t all help appreciated
 HW Helper P: 4,430 Your formula for a is correct. Convert km/h to m/s and find the acceleration. There is a kinematic equation which contains vi,a,t and x. Using that equation find x.To it add safety margin to find the length of the run way.
 P: 18 ok. from what you've said this is what i've tried u=300km/hr=1080m/s v=30km/hr=108m/s t=12s safety margin=250m to get a a=(u-v)/t =(1080-108)/12 =81m/s so: distance covered=u.t + 1/2.a.t^2 =(1080 x 12) + (1/2 x 81 x 144) =12960 + 5832 =18792m then add the safety margin =18792 + 250 = 19042m = 19.042km am I any closer to the answer do you think?
HW Helper
P: 4,430

## finding length of airport runway

Your conversion of km/hr to m/s is wrong.
The plane is slowing down. So the acceleration must be negative.
 P: 18 ok so it would be a=(v-u)/t =(108-1080)/12 =-81m/s so: distance covered=u.t + 1/2.a.t^2 =(1080 x 12) + (1/2 x -81 x 144) =12960 + -360 =12600m then add the safety margin =12600 + 250 = 12850m = 12.85km maybe?
 HW Helper P: 4,430 300km/hr = 300*1000/3600 = ? m/s
 P: 18 had that the first time and thought i was wrong. that being the case so: distance covered=u.t + 1/2.a.t^2 =(83.33 x 12) + (1/2 x -6.25 x 144) =999 + -450 =549m then add the safety margin =549 + 250 = 799m = 0.799km you must hate me by now =]
 HW Helper P: 4,430 Now the answer is correct.
 P: 18 thanks so much rl. you are a legend

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